Maojui

Seccon 2017 - Vigenere3d (Crypto, 100)

2017-12-19

Given : vigenere3d.py

This challenge is Vigenère cipher, but make it to a cube.

Each level has a Vigenère table and rorate 1 alphabet for every floor.

Solve

This challenge needs two key, but they just reverse it.

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main(sys.argv[1], sys.argv[2], sys.argv[2][::-1])

So, I try to decrypt it.

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s = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789abcdefghijklmnopqrstuvwxyz_{}"
t = [[_l((i+j) % len(s), s) for j in range(len(s))] for i in range(len(s))]

def decrypt(c,k) :
    i = 0
    p = ""
    k2 = k[::-1]
    for a in c :
        for level in range(len(s)) :
            if a == t[level][s.find(k[i])][s.find(k2[i])] :
                p += s[level]
        i = (i + 1) % len(k)
    return p

Test every possible key which can get the SECCON{ header.

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cand_khead = ''
cand_ktail = ''
known = 'SECCON{'

for i in range(len(known)) :
    cipher = 'POR4dnyTLHBfwbxAAZhe}}ocZR3Cxcftw9'
    for p in s :
        for q in s :
            k = ( cand_khead + p ).ljust(7,'?')
            k += ( q + cand_ktail ).rjust(7,'?')
            out = decode(cipher,k)
            if out[:i+1] == known[:i+1]:
                cand_khead += p 
                cand_ktail = q + cand_ktail
                print(out)
                print(p,q)
                break
        break

SECCON{Welc0me_to_SECCON_CTF_2017}

Tags: Classical Vigenère
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